Using kernel 2.6.x
How would you script the result below with the following variables using sh (not bash, zsh, etc.) ?
VAR1="abc def ghi"
VAR2="1 2 3"
CONFIG="$1"
for i in $VAR1; do
for j in $VAR2; do
[ "$i" -eq "$j" ] && continue
done
command $VAR1 $VAR2
done
Desired result:
command abc 1
command def 2
command ghi 3
One way to do it:
#! /bin/sh
VAR1="abc def ghi"
VAR2="1 2 3"
fun()
{
set $VAR2
for i in $VAR1; do
echo command "$i" "$1"
shift
done
}
fun
Output:
command abc 1
command def 2
command ghi 3
sh supports both functions and shift.
fun() { ... }. But then set $VAR2 overrides script parameters $1, $2, ..., which is probably not what you want.
A variation on Satō Katsura's answer (here, a self-contained function):
func () {
var=$1
set -- $2
for arg1 in $var; do
printf 'command %s %s\n' "$arg1" "$1" # or cmd "$arg1" "$1" directly
shift
done
}
func "abc def ghi" "1 2 3"
The following would work, but would overwrite the positional parameters of the script that it is in:
var1="abc def ghi"
var2="1 2 3"
set -- $var2
for arg1 in $var1; do
printf 'command %s %s\n' "$arg1" "$1"
shift
done
$1 happens above the non-function solution (like saving $1 in a variable with e.g. CONFIG=$1), it should be ok.
set -- $var2.
sh -x ./test.sh /etc/test2.sh to see what actually happens.
set call. set -- $var2 would overwrite any existing positional parameters, $1, $2, $3 etc.
Following is one of the solution.
#!/bin/sh
var1="a b c"
var2="1 2 3"
set -- $var2
for i in $var1
do
echo $i $1
shift
done
